Problem: The graph of $y=ax^2+bx+c$ is given below, where $a$, $b$, and $c$ are integers.  Find $a$.

[asy]
size(140);
Label f;

f.p=fontsize(4);

xaxis(-3,3,Ticks(f, 1.0));

yaxis(-4,4,Ticks(f, 1.0));

real f(real x)

{

return x^2+2x-1;

}

draw(graph(f,-2.7,.7),linewidth(1),Arrows(6));
[/asy]
The vertex of the parabola appears to be at the value $(-1,-2)$.  Therefore, it is the graph of \[y=a(x+1)^2-2\] for some integer $a$.  We also know that $(0,-1)$ is on the graph, so \[-1=a(0+1)^2-2=a-2.\] Therefore  \[a=\boxed{1}.\]